Fraction Word Problems - Accuplacer Arithmetic

1. Sarah is reading a book. On Monday, she read $\frac{1}{8}$ of the book. On Tuesday, she read $\frac{3}{5}$ of what remained. On Wednesday, she read 84 pages, which completed exactly $\frac{1}{2}$ of what was left after Tuesday. How many pages are in the entire book?
   Answer: 480 pages Let P be total pages. After Monday: $P - \frac{P}{8} = \frac{7P}{8}$ remains. After Tuesday: $\frac{7P}{8} - \frac{3}{5} × \frac{7P}{8} = \frac{7P}{8} × \frac{2}{5} = \frac{7P}{20}$ remains. Wednesday's 84 pages is half of what remained after Tuesday: $84 = \frac{1}{2} × \frac{7P}{20} = \frac{7P}{40}$. Solving: $P = 84 × \frac{40}{7} = 480$ pages. Choice B results from incorrectly calculating Tuesday's remaining fraction. Choice C comes from treating Wednesday's reading as $\frac{1}{3}$ instead of $\frac{1}{2}$ of remaining. Choice D results from calculation errors in the fraction operations.
2. In a class of 60 students, $\frac{2}{5}$ are in the school band. Of the students in the band, $\frac{1}{4}$ play a brass instrument. How many students in the class play a brass instrument?
   Answer: 6 First, find the number of students in the band: $\frac{2}{5} \times 60 = 2 \times 12 = 24$. Next, find the number of these band students who play a brass instrument: $\frac{1}{4} \times 24 = 6$. Therefore, 6 students play a brass instrument.
3. A construction crew completed $\frac{2}{5}$ of a project in the first week and $\frac{1}{4}$ of the remaining work in the second week. What fraction of the entire project remains to be completed?
   Answer: $\frac{9}{20}$ After the first week, $1 - \frac{2}{5} = \frac{3}{5}$ of the project remains. In the second week, they complete $\frac{1}{4}$ of this remaining work: $\frac{1}{4} × \frac{3}{5} = \frac{3}{20}$. Total completed after two weeks: $\frac{2}{5} + \frac{3}{20} = \frac{8}{20} + \frac{3}{20} = \frac{11}{20}$. Therefore, remaining work is $1 - \frac{11}{20} = \frac{9}{20}$. Choice A results from incorrectly calculating $\frac{1}{4} × \frac{2}{5}$ instead of the remaining fraction. Choice C is the amount completed, not remaining. Choice D comes from adding $\frac{2}{5} + \frac{1}{4}$ directly without considering the sequential nature.
4. Two runners are in a race that is 10 kilometers long. Runner A has completed $\frac{3}{4}$ of the race. Runner B has completed $\frac{2}{3}$ of the race. How many more kilometers has Runner A completed than Runner B?
   Answer: $\frac{5}{6}$ First, find the fraction of the race that separates the two runners: $\frac{3}{4} - \frac{2}{3} = \frac{9}{12} - \frac{8}{12} = \frac{1}{12}$. Next, find this fractional distance of the total race length: $\frac{1}{12} \times 10 \text{ km} = \frac{10}{12} \text{ km} = \frac{5}{6}$ km. So, Runner A is $\frac{5}{6}$ of a kilometer ahead of Runner B.
5. On Monday, a painter uses $\frac{1}{3}$ of a can of paint. On Tuesday, she uses $\frac{2}{5}$ of the paint that was remaining in the can. What fraction of the original can of paint is left?
   Answer: $\frac{2}{5}$ After Monday, the remaining paint is $1 - \frac{1}{3} = \frac{2}{3}$ of the can. On Tuesday, she uses $\frac{2}{5}$ of this remainder, which is $\frac{2}{5} \times \frac{2}{3} = \frac{4}{15}$ of the original can. The fraction of paint left is the amount after Monday minus the amount used Tuesday: $\frac{2}{3} - \frac{4}{15} = \frac{10}{15} - \frac{4}{15} = \frac{6}{15} = \frac{2}{5}$.
6. Three friends share a pizza. The first friend eats $\frac{1}{3}$ of the pizza. The second friend eats $\frac{3}{8}$ of what is left. What fraction of the original pizza does the third friend get?
   Answer: $\frac{5}{12}$ After the first friend eats $\frac{1}{3}$, the amount of pizza remaining is $1 - \frac{1}{3} = \frac{2}{3}$. The second friend eats $\frac{3}{8}$ of this remainder, which is $\frac{3}{8} \times \frac{2}{3} = \frac{6}{24} = \frac{1}{4}$ of the original pizza. The total eaten by the first two friends is $\frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12}$. The fraction left for the third friend is $1 - \frac{7}{12} = \frac{5}{12}$.
7. A driver is on a 240-mile trip. The driver covers $\frac{1}{3}$ of the distance before stopping for lunch. After lunch, the driver covers $\frac{3}{8}$ of the remaining distance. How many miles are left to complete the trip?
   Answer: 100 First, calculate the distance covered before lunch: $\frac{1}{3} \times 240 = 80$ miles. The remaining distance is $240 - 80 = 160$ miles. After lunch, the driver covers $\frac{3}{8}$ of this remainder: $\frac{3}{8} \times 160 = 3 \times 20 = 60$ miles. The final remaining distance is $160 - 60 = 100$ miles.
8. After a party, $\frac{1}{8}$ of a pizza is left. Before the last guest took their slices, there was $\frac{1}{2}$ of the pizza remaining. What fraction of the original pizza did the last guest eat?
   Answer: $\frac{3}{8}$ The amount the last guest ate is the difference between the amount of pizza available before they ate and the amount left after they ate. Subtract the final fraction from the starting fraction: $\frac{1}{2} - \frac{1}{8} = \frac{4}{8} - \frac{1}{8} = \frac{3}{8}$. The last guest ate $\frac{3}{8}$ of the original pizza.
9. A library has 120 reference books. It loans out $\frac{1}{4}$ of them. Of the books remaining in the library, $\frac{2}{3}$ are non-fiction. How many non-fiction books are remaining in the library?
   Answer: 60 First, calculate the number of books loaned out: $\frac{1}{4} \times 120 = 30$ books. Then, find the number of books remaining: $120 - 30 = 90$ books. Finally, calculate the number of non-fiction books among those remaining: $\frac{2}{3} \times 90 = 2 \times 30 = 60$. There are 60 non-fiction books remaining.
10. The perimeter of a rectangular garden is $25 \frac{1}{2}$ feet. If the length of the garden is $8 \frac{1}{4}$ feet, what is the width of the garden in feet?
    Answer: $4 \frac{1}{2}$ The perimeter of a rectangle is $2L + 2W = P$. The combined length of the two long sides is $2 \times 8 \frac{1}{4} = 2 \times \frac{33}{4} = \frac{66}{4} = 16 \frac{1}{2}$ feet. Subtract this from the total perimeter to find the combined length of the two widths: $25 \frac{1}{2} - 16 \frac{1}{2} = 9$ feet. Since this is the length of two widths, divide by 2 to find the length of one width: $9 \div 2 = 4 \frac{1}{2}$ feet.
11. A fuel tank was $\frac{7}{8}$ full. After a trip, the tank is now $\frac{1}{4}$ full. If the tank holds 24 gallons when full, how many gallons of fuel were used on the trip?
    Answer: 15 First, find the fraction of the tank that was used: $\frac{7}{8} - \frac{1}{4} = \frac{7}{8} - \frac{2}{8} = \frac{5}{8}$. Next, find this fractional part of the total capacity: $\frac{5}{8} \times 24 = 5 \times \frac{24}{8} = 5 \times 3 = 15$. Therefore, 15 gallons were used.
12. A carpenter has a board of wood that is 15 feet long. He needs to cut as many shelves as possible that are each $2 \frac{1}{3}$ feet long. After cutting the maximum number of whole shelves, how much wood, in feet, will be left over?
    Answer: 1 First, determine how many shelves can be cut by dividing the total length by the shelf length: $15 \div 2 \frac{1}{3} = 15 \div \frac{7}{3} = 15 \times \frac{3}{7} = \frac{45}{7} = 6 \frac{3}{7}$. This means 6 whole shelves can be cut. Calculate the amount of wood used for these 6 shelves: $6 \times 2 \frac{1}{3} = 6 \times \frac{7}{3} = 14$ feet. The leftover wood is the original length minus the used length: $15 - 14 = 1$ foot.
13. Three partners, Alex, Beth, and Chandra, invest in a business. Alex invests $\frac{1}{4}$ of the total capital, and Beth invests $\frac{2}{5}$ of the total capital. What fraction of the capital must Chandra invest?
    Answer: $\frac{7}{20}$ First, find the total fraction of capital invested by Alex and Beth combined. Find a common denominator for the fractions: $\frac{1}{4} + \frac{2}{5} = \frac{5}{20} + \frac{8}{20} = \frac{13}{20}$. The total capital is represented by 1. To find Chandra's share, subtract the combined share of Alex and Beth from 1: $1 - \frac{13}{20} = \frac{20}{20} - \frac{13}{20} = \frac{7}{20}$.
14. A delivery truck has a weight capacity of $2 \frac{1}{2}$ tons. If it is being loaded with crates that each weigh $\frac{1}{8}$ of a ton, what is the maximum number of crates that can be loaded onto the truck?
    Answer: 20 To find the number of crates, divide the total capacity of the truck by the weight of a single crate. First, convert the mixed number to an improper fraction: $2 \frac{1}{2} = \frac{5}{2}$. Now, perform the division: $\frac{5}{2} \div \frac{1}{8} = \frac{5}{2} \times \frac{8}{1} = \frac{40}{2} = 20$. The truck can hold a maximum of 20 crates.
15. A farmer has a 50-pound bag of grain. On Monday, he uses $\frac{1}{5}$ of the bag. On Tuesday, he adds 10 more pounds of grain to the bag. On Wednesday, he uses $\frac{1}{3}$ of the new total amount in the bag. How many pounds of grain are left?
    Answer: $33 \frac{1}{3}$ On Monday, the farmer uses $\frac{1}{5} \times 50 = 10$ pounds. The amount remaining is $50 - 10 = 40$ pounds. On Tuesday, he adds 10 pounds, so the new total is $40 + 10 = 50$ pounds. On Wednesday, he uses $\frac{1}{3}$ of this amount: $\frac{1}{3} \times 50 = \frac{50}{3} = 16 \frac{2}{3}$ pounds. The amount left is $50 - 16 \frac{2}{3} = 33 \frac{1}{3}$ pounds.
16. A water tank is $\frac{3}{4}$ full. After using $\frac{1}{6}$ of the water currently in the tank for irrigation, and then adding 45 gallons, the tank becomes $\frac{5}{6}$ full. What is the total capacity of the tank?
    Answer: 216 gallons Let C be the tank's capacity. Initially: $\frac{3C}{4}$ gallons. After using $\frac{1}{6}$ of current water: $\frac{3C}{4} - \frac{1}{6} × \frac{3C}{4} = \frac{3C}{4}(1 - \frac{1}{6}) = \frac{3C}{4} × \frac{5}{6} = \frac{5C}{8}$. After adding 45 gallons: $\frac{5C}{8} + 45 = \frac{5C}{6}$. Solving: $45 = \frac{5C}{6} - \frac{5C}{8} = \frac{20C - 15C}{24} = \frac{5C}{24}$. Therefore: $C = 45 × \frac{24}{5} = 216$ gallons. Choice A comes from setting up the equation incorrectly. Choice C results from using $\frac{1}{6}$ of total capacity instead of current water. Choice D comes from computational errors in fraction arithmetic.
17. A container is filled with $6 \frac{1}{2}$ gallons of water. Another $3 \frac{1}{3}$ gallons are added. If the container's total capacity is 15 gallons, what fraction of the container is now full?
    Answer: $\frac{59}{90}$ First, find the total amount of water in the container by adding the two amounts: $6 \frac{1}{2} + 3 \frac{1}{3} = (6+3) + (\frac{1}{2} + \frac{1}{3}) = 9 + (\frac{3}{6} + \frac{2}{6}) = 9 \frac{5}{6}$ gallons. Convert this to an improper fraction: $9 \frac{5}{6} = \frac{59}{6}$. To find the fraction of the container that is full, divide the amount of water by the total capacity: $\frac{59}{6} \div 15 = \frac{59}{6} \times \frac{1}{15} = \frac{59}{90}$.
18. A baker made a batch of dough weighing 40 pounds. He used $\frac{3}{5}$ of the dough to make loaves of bread. He then used $\frac{1}{4}$ of the remaining dough to make dinner rolls. How many pounds of dough were left over?
    Answer: 12 First, calculate the amount of dough used for bread: $\frac{3}{5} \times 40 = 24$ pounds. Then, find the amount of dough remaining: $40 - 24 = 16$ pounds. Next, calculate the amount of dough used for dinner rolls from the remainder: $\frac{1}{4} \times 16 = 4$ pounds. Finally, subtract the dough used for rolls from the remainder to find the final amount left over: $16 - 4 = 12$ pounds.
19. A company's profit was $64,000 in one year. The next year, the profit was $\frac{7}{8}$ of the previous year's profit. In the third year, the profit was again $\frac{7}{8}$ of the second year's profit. What was the company's profit in the third year?
    Answer: $49,000 To find the profit in the second year, calculate $\frac{7}{8}$ of $64,000: $\frac{7}{8} \times 64000 = 7 \times 8000 = \$56,000$. To find the profit in the third year, calculate $\frac{7}{8}$ of the second year's profit: $\frac{7}{8} \times 56000 = 7 \times 7000 = \$49,000$.
20. A recipe that yields 12 cookies requires $1 \frac{3}{4}$ cups of sugar. How many cups of sugar are needed to make 30 cookies?
    Answer: $4 \frac{3}{8}$ First, find the scaling factor for the recipe: $\frac{30}{12} = \frac{5}{2}$. Convert the sugar amount to an improper fraction: $1 \frac{3}{4} = \frac{7}{4}$. Multiply the sugar amount by the scaling factor: $\frac{7}{4} \times \frac{5}{2} = \frac{35}{8}$. Convert back to a mixed number: $35 \div 8 = 4$ with a remainder of 3, so the answer is $4 \frac{3}{8}$ cups.

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